PWS 340 Final

Sickle-cell anemia is a human genetic disease arising from a mutation in the gene (HBB) encoding the hemoglobin B protein. Individuals homozygous for mutant alleles have sickle-cell anemia but individuals who are heterozygous are resistant to malaria. A mutant HBB allele, designated S, associated with sickle-cell disease is found in some parts of Africa. The wild-type allele is referred to as A. The Niokholonko people are subsistence farmers in Eastern Senegal. In 1975, Mauran-Sendrail and Bouloux (Annals of Human Biology 2:129-136) reported that from a study of 599 Niokholonko people who represented 80% of the entire population, 483 had the genotype HBB*A/*A, 114 had the genotype HBB*A/*S and two had the genotype HBB*S/*S. What is the F coefficient for this population, with respect to HBB?

-0.0717

Achondroplasia is a dominant genetic disorder that results in dwarfism. In those with the condition, the arms and legs are short, while the torso is typically of normal length. Those affected have an average adult height of 4 ft 4 in for males and 4 ft for females. In 80% of cases Achondroplasia is caused by a mutation in FGFR3 gene. This gene is mainly responsible for making the protein, fibroblast growth factor receptor 3. This protein contributes to the production of collagen and other structural components in tissues and bones. When the FGFR3 gene is mutated, it interferes with how this protein interacts with growth factors leading to complications with bone production. Cartilage is not able to fully develop into bone, causing the individual to be disproportionately shorter in height. On average four babies are born with achondroplasia for every 50,000 babies born to parents without achondroplasia. Use this infor

0.00004 mutations

Rudy Giuliani was mayor of New York from 1994 until 2001. He has been married three times. His first marriage was to Regina Peruggi in 1968. The marriage did not last, and they had it annulled by the Catholic Church because, according to Giuliani, he discovered that Regina was his second cousin. As it turns out, Peruggi was Giuliani's second cousin, once removed. Giuliani and Peruggi did not have any children together but if they had, what would the inbreeding coefficient have been? You might need to refer to the English Kinship System slide provided in the following slide.

0.0078125

Genotype frequencies for the HBB gene in a Bedik population from eastern Senegal were published by Mauran-Sendrail and Bouloux (Annals of Human Biology 2:129-136). They give the values of the selection coefficients, s and t, as 0.1659 and 1, respectively. What value of q results in balanced polymorphism of the HBB gene for the Bedik population? (Hint: What will Δq be when balance is reached?) A.0.1254 C. 0.3475 E. 0.8577 0.1423D.0.5000

0.02025

The MN gene (OMIM 111300) is located on chromosome 4 in humans. It encodes a polypeptide that is modified into a glycoprotein and deposited on the surface of red blood cells. Two alleles, MN*M and MN*N are common in human populations. The MN blood type of an individual can be readily determined with an immunological assay similar to the test used to determine ABO blood type. The M and N alleles are codominant such that type M (MN*M/*M), type N (MN*N/*N) and type MN (MN*M/*N) blood can be easily distinguished. In 1980, Neel et al. (Ann. Hum. Genet. 44:37-54) reported the following blood types in 1747 people from the Ticuna tribe from the Amazon rain forest of Brazil. Type M 1409 Type MN 310 Type N 28 What is the inbreeding coefficient (F) for the MN gene in this population?

0.0539

What is the probability of having all four children be boys?

0.0625

In Drosophila melanogaster, the vermilion phenotype is bright-red eyes and is the result of the recessive mutant allele v. The sable phenotype is black body color and is caused by the recessive allele s. In 1916, Morgan and Bridges (Carnegie Inst. Publ. 327) reported the results of an experiment in which they crossed a vermilion female with a sable male. All of the F1 females were wild type and all of the F1 males were vermilion. They allowed the F1 females and males to mate with one another to produce an F2 generation. The following lists male flies only in the F2 generation. wild type 15 vermilion 101 sable 74 vermilion, sable 9 Total: 199 What is the frequency of recombination between the two genes (do not use centiMorgans; your answer must be between 0 and 1)

0.120

Dark and light coat color in many animals is determined by alleles of the Mc1r gene. Pocket mice have two alleles of this gene, D and d, which generate dark and light coats, respectively, with D completely dominant to d. In 2004, Hoekstra et al. (Evolution 58:1329-1341) measured relative fitness values for pocket mice residing on dark lava sites in southern Arizona. From their experiments they determined that w = 1 for dark-coated mice and w = 0.61 for light-coated mice. On one test site they counted 44 DD mice, 10 Dd mice and 3 dd mice. What is the frequency of the d allele in this population of mice? What is the selection coefficient for the dd genotype? What would be the change in frequency of the d allele after one generation for these mice?

0.14 0.39 -0.0066553

In 1983, Le Pendu et al. (American Journal of Human Genetics 35:484-496) published a pedigree of a human family in which the father had the genotype ABO*B/O; FUT1*H/O, and the mother had the genotype ABO*O/O; FUT1*H/O. The ABO and FUT1 loci are not linked. The FUT1*O allele is non-functional. Is this truly a dihybrid cross or something new? They had nine children, five with B antigen in their blood (type B blood), and four without B antigen in their blood (type O blood or the Bombay phenotype). Given the genotypes of the parents, write in the answer box the probability of this particular outcome (5 children with the B antigen, and 4 without, in a family of 9). What fraction will have the B antigen?

0.14257

The testing of nuclear bombs in the 1950s and 1960s in the United States released large amounts of the radioisotope strontium-90 into the atmosphere. Despite assurances that what went up into the stratosphere stayed in the stratosphere, it was soon discovered that the strontium made its way back to earth rather quickly, most likely via rainwater. In the US, much of the strontium ended up in the wheat fields of the Great Plains states where giant storms reached high into the atmosphere. One experiment was designed to measure the ability of different wheat varieties to take up strontium from the soil to ascertain the genetic contribution of the trait. The data below is taken from that experiment. What is the estimated number of genes involved in uptake of strontium in wheat? No. plants Mean Strontium (ppm) Variance (ppm2) Kenya 117A 28 161 334 Kentana 52 29 110 234 F1 26 159 397 F2 81 158 1083

0.42

In 1983, Le Pendu et al. (American Journal of Human Genetics 35:484-496) published a pedigree of a human family in which the father had the genotype ABO*B/O; FUT1*H/O, and the mother had the genotype ABO*O/O; FUT1*H/O. (The ABO and FUT1 loci are not linked.) The couple had nine children, five with B antigen in their blood (type B blood), three with type O blood and one with the Bombay phenotype. Given the probabilities already determined for each phenotype, what is the overall probability of this particular outcome (5 children with the B antigen, 3 with type O blood and 1 with the Bombay phenotype, in a family of 9)?

0.4927

The rodenticide warfarin was originally introduced during World War II. Although it was initially very successful, its effectiveness has lessened over time because of an increase in resistance among some target populations of rats. Among Norway rats in Great Britain, resistance results from an otherwise harmful mutation R in a gene in which the normal sensitive allele is denoted S. In 1985, R.M. May (Nature 315:12-13) calculated the relative fitnesses of the SS, SR, and RR genotypes in the absence of warfarin to be 1.00, 0.77 and 0.46, respectively. The reduced fitness of the RR genotype(in the absence of warfarin) appears to result from an excessive requirement for vitamin K. •In the absence of warfarin, what is the value of the selection coefficient s for this population of rats? •If the frequency of the R allele (q0) in a population of Norway rats is initially 0.34, what will be its frequency (q1) after a s

0.54 0.27

Tatuo Aida investigated the genetic basis of color variation in the Medaka (Aplocheilus latipes), a small fish found naturally in Japan (T. Aida. 1921. Genetics 6:554-573). Aida found that genes at two loci (B, b and R, r) determine the color of the fish: fish with a dominant allele at bot loci (B_R_) are brown, fish with a dominant allele at the B locus only (B_rr) are blue, fish with a dominant allele at the R locus only (bbR_) are red, and fish with recessive alleles at both loci (bb rr) are white. Aida crossed a homozygous brown fish with a homozygous white fish. He then backcrossed the F1 with the homozygous white parent and obtained 228 brown fish, 230 blue fish, 237 red fish, and 222 white fish. What results would you expect if you crossed a homozygous red fish with a homozygous blue fish and then backcrossed the F1 with a homozygous red parental fish?

1 Brown : 1 Red

Polydactyly is a human autosomal dominant condition that produces extra fingers and toes. Studies of hundreds of families with polydactyly have determined that penetrance for the dominant allele is 70%. Hospital based survey of live births finds that 25 in 1,000,000 infants has a new case of polydactyly. Use this information to estimate the mutation rate of this gene.

1.786 x 10^-5

Eugenics is the term employed for the selective breeding of humans to bring about improvements in populations. Although it was once practiced in the United States, it is now considered to be repugnant as well as scientifically unsound. As a eugenic measure, it has been suggested that individuals suffering from supposed genetic disorders should be prevented (sometimes by law) from reproducing (by sterilization, if necessary) in order to reduce the frequency of the disorder in future generations. Suppose that such a recessive trait were present in the population at a frequency of 1 in 40,000 and that affected individuals were prohibited from reproducing. In 10 generations, or about 250 years, what would be the frequency of the condition? (Assume all Hardy-Weinberg assumptions except for selection)

1/44,100

Later, Morgan reported the results of an experiment involving D. melanogaster genes that were not sex-linked. He crossed a male fly that was homozygous for purple eyes (pr pr) and vestigial wings (vg vg) with a wild-type female fly. All of the F1 progeny were wild type. He backcrossed the F1 females to the homozygous mutant males to obtained the following backcross progeny wild type 1339 purple, vestigial 1195 purple 154 vestigial 151 Calculate the recombination frequency between purple-eye and vestigial and correct for double crossovers using the Kosambi function. What is the linkage distance in cM between the two genes?

10.91 cM 10.87 cM

The Ac/Ds DNA transposable element system of maize was first discovered by Barbara McClintock, leading to her 1983 Nobel Prize in Medicine. The Ac Activator element is autonomous, whereas the Ds Dissociation element requires an Activator element to transpose. In 1950, McClintock (PNAS 36:344-355) described experiments in which she hybridized an inbred line of maize with colorless kernels due to the c-m1 allele (this allele carries a Ds element; the "m" stands for "mutable"), with a second inbred line with colored kernels that was homozygous for the presence of an Ac element. Assuming that the color gene and the Ac elements assort independently, what is the expected ratio of phenotypes in the F2 generation?

12 colored, 3 spotted, 1 colorless

In Drosophila melanogaster, black body (b) is recessive to gray body (b+), purple eyes (pr) are recessive to red eyes (pr+), and vestigial wings (vg) are recessive to normal wings (vg+). These three genes are linked with pr located between b and vg at a genetic distance of 6 cM from b and 13 cM from vg. Previous experiments have shown that the interference associated with these three genes is 0.5. A fly with a black body, purple eyes and vestigial wings is crossed with a fly with a gray body, red eyes and normal wings. All of their F1 progeny are wild-type. The F1 females are mated with male flies that have black bodies, purple eyes and vestigial wings. Ten thousand progeny are produced from this testcross. How many of the testcross progeny are expected to have: 1) Purple eyes (pr) but be wild-type for body color (+) and wing (+) type?

19.5

Mendel conducted monohybrid crosses for all seven traits he included in his experimental pea plants. Based on the results of these crosses, he predicted that in the F2 generation, the ratio of plants with a dominant phenotype to those with a recessive phenotype is 3:1. He further postulated that the F2 genotypic ratio is 1:2:1, that is one-fourth of the plants are homozygous for dominant alleles, one-half are heterozygous and one-fourth are homozygous for recessive alleles. To test this hypothesis, Mendel performed test crosses, pollinating the dominant F2 plants with pollen from homozygous recessive plants. He observed that some of the plants yielded test-cross progeny that were all dominant, whereas others produced a mixture of dominant and recessive progeny. In those test crosses that yielded a mixture, what do you expect the ratio of dominant to recessive to be?

1:1

Mendel conducted monohybrid crosses for all seven traits he included in his experimental pea plants. Based on the results of these crosses, he predicted that in the F2 generation, the ratio of plants with a dominant phenotype to those with a recessive phenotype is 3:1. He further postulated that the F2 genotypic ratio is 1:2:1, that is one-fourth of the plants are homozygous for dominant alleles, one-half are heterozygous and one-fourth are homozygous for recessive alleles. To test this hypothesis, Mendel self-pollinated dominant F2 plants and observed that some of the plants yielded progeny that were all dominant, whereas others produced progeny that segregated 3:1 dominant to recessive. Suppose Mendel planted 60 dominant F2 plants and allowed them to self pollinate. How many of the 60 plants would you expect to produce only dominant offspring?

20 (1/3 of the total)

•Hitler reasoned that alleles that caused recessive genetic disorders could be removed from the population by sterilizing afflicted individuals. •Suppose that the frequency of a particular autosomal recessive disorder is 1/5000 people. -How many years (assume 25 years/generation) would be required to reduce the frequency to 1/10,000 using Hitler's eugenic philosophy?

29.28 generations or 732 years

Arabidopsis thaliana is a model organism used to study the genetics and development of plants. In 2005, Brukhin et al. (Plant Cell 17:2723-2737) described lethal alleles of the A. thaliana gene RPN1a, which encodes a subunit of the 26S proteasome. In one experiment they self-pollinated plants heterozygous for the rpn1a-1 lethal allele and observed seed set in the open siliques (fruits of the Arabidopsis plant). In the figure below, the top silique is from a homozygous wild-type plant, whereas the bottom silique is from a self-pollinated heterozygote with the arrows indicating aborted seeds. Brukhin et al. counted 1016 seeds from self-pollinated heterozygotes; they identified 772 normal and 244 aborted seeds. What ratio do you expect?

3 alive: 1 dead

If 9% of an African population is born with a severe form of sickle-cell anemia (SS) and assuming HWE, what percentage of the population will be more resistant to malaria because they are heterozygous (AS) for the sickle-cell gene?

42%

Two heterozygous people marry and plan on having four children, what is the probability that at least one child will have PKU? PKU (phenylketonuria) is caused by a recessive allele. •Probability of having a child with PKU is ¼ •Probability of having a carrier or homozygous dominant child is ¾

68.36%

How many generation are required to reduce the frequency of albino rabbits in a population from 25% to 1% (albinism is controlled by a recessive allele)?

8

Sickle-cell anemia is a human genetic disease arising from a mutation in the gene (HBB) encoding the hemoglobin B protein. Individuals homozygous for mutant alleles have sickle-cell anemia but individuals who are heterozygous are resistant to malaria. A mutant HBB allele, designated S, associated with sickle-cell disease is found in some parts of Africa. The wild-type allele is referred to as A. The Niokholonko people are subsistence farmers in Eastern Senegal. In 1975, Mauran-Sendrail and Bouloux (Annals of Human Biology 2:129-136) reported that from a study of 599 Niokholonko people who represented nearly the entire population, 483 had the genotype HBB*A/A, 114 had the genotype HBB*A/S and two (2) had the genotype HBB*S/S. What are the frequencies of the A and S alleles?

A - 0.9015 S - 0.09849

Bobbed (bb) is a pseudoautosomal recessive trait in Drosophila melanogaster that is characterized by shorter and thinner than normal bristles on the back of the head. If a bobbed male is mated with an attached X (X^XY) homozygous, wild-type female, what is the expected proportion of bobbed flies in the adult progeny? Please note that the X chromosome is necessary for survival. A.0 B. 1/4 C. 1/2 D. 3/4 E. All bobbed males

A. 0

In an isolated population of 50 desert bighorn sheep, a mutant recessive allele c has been found to cause curled coats in both males and females. The normal dominant allele C produces straight coats. A biologist studying these sheep counts four (4) with curled coats and takes blood samples for DNA marker analysis, which reveals that 17 of the straight-coated sheep are carriers of the c allele. What is the inbreeding coefficient F for this population? A. 0.1 B. 0.2 C. 0.3 D. 0.4 E. 0.5

A. 0.093

Albinism is an inherited condition in humans in which pigment is not produced in the hair, skin or iris. Mutant alleles of the gene responsible for albinism are recessive. Dr. Coleman's wife has an uncle that was albino. He married a woman who was not albino. They had six children of which only one was not albino. What is the probability of this outcome? A. 0.09375 B. 0.3560 C. 0.015625 D. 0.004395 E. 0.000732

A. 0.09375

Semi-sterile maize nonvirescent (wildtype) plants heterozygous for a reciprocal translocation between chromosomes 1 and 2 were crossed with chromosomally normal plants homozygous for a recessive allele in chromosome 2 that results in virescent leaves (Figure 1, see below). F1 plants were produced that were both semisterile and fully fertile. When semisterile F1 plants were testcrossed to fully-fertile virescent plants, the following data were obtained: semisterile, virescent 39 fertile, virescent 231 semisterile, nonvirescent 287 fertile, nonvirescent 31 What is the recombination frequency between the virescent gene locus and the translocation break point, rounded to one decimal place? Assume duplicate deficient gametes are not viable. A) 0.1 B) 0.2 C) 0.3 D) 0.4 E) 0.5

A. 0.1

In maize trisomics, n+1 pollen is not viable. A dominant allele at the B locus in the maize genome produces purple color instead of the recessive phenotype bronze. If a Bbb trisomic plant is pollinated by a BBb plant, what proportion of the progeny produced will be trisomic and have a bronze phenotype? A. 1/18 B. 1/6 C. 1/4 D. 1/3 E. 1/2

A. 1/18

In 1938, Marcus Rhoades (Genetics 23:377-397) reported the first known study demonstrating the inheritance patterns associated with a transposon. He obtained an ear of corn from another researcher that had colored, spotted and colorless kernels (see picture below). Colored seeds had at least one dominant allele at the C1 locus whereas the colorless kernels had a homozygous c1c1 genotype. Rhoades hypothesized that the spotted phenotype arose from the presence of a dominant Ct allele that interacted with c1 alleles to revert them to wild-type C1 alleles in some but not all cells of the seed coat. Recessive ct alleles would have no effect on c1 alleles. Rhoades ear of corn came from the self-pollination of a plant that was heterozygous for the Ac element (Ac+Ac-) and heterozygous for the Ds element mutation leading to the cm-1 allele. Assuming that the color gene and the Ac elements assort independently, what ratio of

A. 12:3:1

In chickens, barred plumage (white bars on black feathers) is under control of a dominant gene B, located on the Z chromosome. The recessive b allele produces nonbarred plumage. What phenotypic ratio do you expect in the F1 progeny of a cross between a barred hen and a nonbarred rooster? A.1:1 barred males to nonbarred females B.3:1 barred males to nonbarred females C.1:1 nonbarred males to barred females D.3:1 nonbarred males to barred females E.All barred

A. 1:1 barred males to nonbarred females

In maize a recessive allele c confers a colorless kernel when homozygous and the recessive allele wx confers a kernel with a waxy texture when homozygous. The dominant allele C gives a colored kernel and the dominant allele Wx gives a starchy texture to the kernel. In 1918, Bregger (American Naturalist 52:57-61) reported data from a cross between a plant grown from a homozygous colored, starchy kernel (CC WxWx), and a plant grown from a homozygous colorless, waxy kernel (cc wxwx). The F1 kernels were all doubly heterozygous (Cc Wxwx) and were colored and starchy. After self-fertilizing the F1 plants, Bregger recorded the following data in the F2 generation: colored, starchy 1774 colorless, starchy 279 colored, waxy 263 colorless, waxy 420 Under the hypothesis that kernel color and starch alleles assort independently, what is the expected ratio of colored/starchy kernels in this F2 population? A.9/16 C. 1/4 E. 1/16

A. 9/16

In 2001, researchers at Texas A&M reported the successful cloning of a house cat which they named CC (carbon copy). The cloning was done by somatic cell nuclear transfer, taking a donor nucleus from an ovarian somatic cell of a calico cat named Rainbow and inserting it into an enucleated egg cell. After the embryo began to develop in vitro, they implanted it in the uterus of a surrogate mother, a tabby cat. Considering the Lyon (Barr body) hypothesis, what phenotype would you predict for CC? A. Black and white or orange and white B. Calico (orange, black and white) C. Tabby (striped pattern)

A. Black and white or orange and white

Duchenne (DMD) and Becker (BMD) muscular dystrophies are genetic diseases arising from mutations in the dystrophin gene located on the human X chromosome. The symptoms associated with BMD are much milder than those exhibited by people with DMD. About 2/3 of all mutations associated with both types of muscular dystrophy are deletions within the very large (2.4 Mb, 78 exons) dystrophin gene. Remarkably, the size of the deletion is not correlated to the severity of the disease. In 2007 van Deutekom et al. (N. Engl. J. Med. 357:2677-2688) reported a DMD patient with a deletion spanning exons 48 - 50, whereas in 2010, Helderman-van den Eneden et al. (Neuromuscular Dis. 20:251-254) reported a BMD patient with a deletion spanning exons 45 - 51. In the figure below, the top sequence shows the 3' end of exon 47, the 5' end of exon 51 and the deleted exons/introns in between as described for the DMD patient. The bottom sequenc

A. Frameshift

Rabbits may either have short hair or long hair dependent on their genotype at a single genetic locus (S). The following crosses were performed with the results shown for each: Parents Progeny short x short 4 short and 3 long short x short 8 short short x long 10 short and 3 long short x long 3 short and 1 long long x long 7 long Which is the dominant trait? A.Short hair B.Long hair

A. Short hair

The data below, excerpted from a paper by Heui-Soo Kim and Osamu Takenaka, are short DNA sequences from five species: human, chimpanzee, gorilla, orangutan, and baboon. The sequences, each 50 bases long, represent a short piece of the SRY gene. The complete human sequence is given. The sequences for the other four species are shown only where they differ from the human sequence. Calculate the genetic difference between each pair of species. For example, chimpanzees and gorillas differ in 2 out of 50 bases, or 4 percent. Use UPGMA to reconstruct the phylogeny for the five species listed based on this short sequence.

A. in notes

Woman lacked both A and B antigens and was typed "O". But based on her pedigree her mother was a type AB and her father an AO. Hence, she genetically must have been a A, AB, or B, but was functionally an O type! HOW? Moreover, she had children that were functionally type "B", where she was the obvious donor of the B allele - hence her B allele was functional! •Indeed, she had inherited a rare recessive mutation, FUT1*O, which prevented her from synthesizing the complete H antigen. •Without the H antigen (lack of the fucose ring), the ABO*A and the ABO*B alleles don't have a substrate to work on. We designate her blood type as Oh What is her full genotype at the ABO locus? What type of blood should be given to this person (what antigen does she have: A, B, H or something else)?

AB or BO Blood from another individual without any antigen, including the O antigen

In modern U.S. agriculture, nearly all corn is hybrid corn. A major system for producing hybrid corn is cytoplasmic male sterility (CMS). Some lines of corn carry a dominant restorer gene (R) that can restore pollen fertility in male-sterile plants. 1. If a male-sterile plant is crossed with pollen from a plant homozygous for a nuclear restorer gene (R) what would be the phenotype of the F1 plant? (Sterile or Fertile)? 2. If the F1 plants were used as females in a testcross with pollen from a homozygous non-restorer plant what would be the frequencies of the male-fertile and male-sterile plants and what would be the cytoplasmic type (mf or ms) of each group?

All fertile 50 % sterile; 50% fertile; all ms mitochondria

In the domestic cat, black coat color is caused by the o allele and orange coat color is caused by the O allele at the same locus. In 1964, N.B. Todd (Heredity 19:47-51) reported that among stray cats sampled in Boston, 102 females and 99 males were black, 4 females and 28 males were orange, and 48 females and no males were tortoiseshell What is the frequency of the O allele in this population of stray cats? (Hint: Use all the cats to determine this, male and females). A. 0.1495 B. 0.1931 C. 0.1993 D. 0.1860 D. 0.5336

B. 0.1931

The mean and variance of plant height of two highly inbred strains (P1 and P2) and their progeny (F1 and F2) are shown in the table below. Strain Mean Variance P1 34.2 4.2 P2 55.3 3.8 F1 44.2 5.6 F2 46.3 10.3 What is the broad-sense heritability for height in this species of plant (rounded to the nearest whole number)? A. 0.46 B. 0.56 C. 0.59 D. 0.63 E. 0.76

B. 0.56

In 1923, Bridges and Morgan reported the following data from a three-factor linkage testcross involving chromosome 3 in Drosophila melanogaster. The three loci are ca (claret), e (ebony), and ro (rough). All three mutant phenotypes are due to recessive alleles. Wild type 49 Ebony 1 Claret 395 Rough 119 Ebony, rough 370 Ebony, claret 89 Rough, claret 1 Ebony, rough, claret 66 What is the linkage distance between rough and claret, in cM? A.10.6 B. 10.7 C. 14.4 D. 19.3 E. 30.0 What is the interference calculated from this data set? A. 0.1 B. 0.3 C. 0.5 D. 0.7 E. 0.9

B. 10.7 E. 0.9

In their 1916 publication on sex-linked inheritance in Drosophila, Morgan and Bridges reported linkage between club wing (cl), vermillion eyes (v) and cherry eyes (wc). They crossed a homozygous club, vermillion, cherry female with a wild-type male and interbred the F1 flies to produce an F2 generation. The observed the following phenotypes in the F2 flies. Wild-type 471 Cherry, club, vermillion 356 Cherry 83 Club, vermillion 65 Cherry, club 71 Vermillion 111 Cherry, vermillion 5 Club 9 Are the alleles for cherry, club and vermillion in coupling or repulsion? What is the order of the three genes on the X chromosome? What is the genetic linkage distance, in cM, between cherry and club? A.12.6 B. 13.8 C. 16.2 D. 16.7 E. 17.8

B. 13.8

A species of wheat (Triticum dicoccum) has a chromosome number of 28, while a species of rye (Secale cereale) has a chromosome number of 14. Sterile hybrids can be produced by crossing these two plants. What would be the expected chromosome number in the somatic cells of the hybrids? A.14 B.21 C.28 D.35 E.42

B. 21

How many chromatids are present in a normal human cell in prophase II of meiosis? A. 23 B. 46 C. 69 D. 92 E. 138

B. 46

A horse has 64 chromosomes and a donkey has 62 chromosomes. A cross between a female horse and a male donkey produces a mule, which is usually sterile. How many chromosomes does a mule have? A.62 B.63 C.64 D.124 E.126

B. 63

Many of the experiments with transposons in maize were conducted with mutable alleles that are expressed in the aleurone layer of the endosperm. The c-m1 allele contains a Ds element inserted into the C1 locus and causes kernels to be light colored when homozygous. Wild-type C1 alleles produce a purple-colored kernel. What is the expected phenotype of the progeny kernels if a plant homozygous for the C1 (C1/C1) allele and for an independent Ac element (Ac+/Ac+) is crossed with a light colored (non-spotted) plant homozygous for c-m1? A.All light-colored B.All purple-colored C.3:1 purple:light D.1:1 purple:spotted E.All spotted

B. All purple-colored

The MN blood group system is under the control of an autosomal locus found on chromosome 4, with two alleles designated LM and LN. The blood type is due to a glycoprotein present on the surface of red blood cells, which behaves as a native antigen. Phenotypic expression at this locus is codominant because an individual may exhibit either one or both antigenic substances. A woman has blood type A M. She has a child with blood type AB MN. Which of the following men cannot be excluded as the father of the child? A.Josh - Type O N B.Spencer - Type AB MN C.Sterling - Type B MN D.Alex - Type A N E.Thomas - Type AB M

B. Spencer C. Sterling

A female fruit fly with orange eyes and long wings is mated with a male fly with red eyes and short wings. The F1 females have red eyes and long wings; the F1 males have orange eyes and long wings. The F1 flies are crossed to yield the following F2 progeny, with no difference between the sexes. 47 long wings, red eyes 45 long wings, orange eyes 17 short wings, red eyes 14 short wings, orange eyes Which of the following is true? A.Both wing length and eye color are autosomal B.Wing length is autosomal, eye color is sex-linked C.Wing length is sex-linked, eye color is autosomal D.Both wing length and eye color are sex-linked

B. Wing length is autosomal, eye color is sex-linked

Tatuo Aida investigated the genetic basis of color variation in the Medaka (Aplocheilus latipes), a small fish found naturally in Japan (T. Aida. 1921. Genetics 6:554-573). Aida found that genes at two loci (B, b and R, r) determine the color of the fish: fish with a dominant allele at bot loci (B_R_) are brown, fish with a dominant allele at the B locus only (B_rr) are blue, fish with a dominant allele at the R locus only (bbR_) are red, and fish with recessive alleles at both loci (bb rr) are white. Aida crossed a homozygous brown fish with a homozygous white fish. He then backcrossed the F1 with the homozygous white parent and obtained 228 brown fish, 230 blue fish, 237 red fish, and 222 white fish. What are the genotypes of the backcross progeny?

BbRr, Bbrr, bbRr, bbrr

In 1987, Marsden et al. (Genetics 116:299-311) reported on patterns of inheritance in the autotetraploid Cope's gray tree frog (Hyla versicolor). To conduct their analysis they used genetic markers. For one of their markers, Mpi, they identified two alleles, which they designated Mpi-1 and Mpi-2. They crossed a female frog duplex for Mpi-1 (1122) with a male frog triplex for Mpi-1 (1112). In the progeny of this cross they detected the following genotypes: 1111 8 1112 61 1122 43 1222 8 What proportion of the progeny do you predict to have the genotype 1222 ? A. 1/16 B. 1/15 C. 1/12 D. 1/6 E. 1/4

C. 1/12

In 2006, Hatakeyama et al. (Hum. Reprod. 21:976-979) submitted a case study of a male heterozygous for a Robertsonian translocation between chromosomes 13 and 21. They used FISH to analyze meiotic segregation patterns in the sperm of the study subject with the following results: Balanced normal or translocation 9036 Nullisomy 21 387 Disomy 21 208 Nullisomy 13 399 Disomy 13 139 Based on theoretical considerations, what proportion of the man's sperm do you expect to be balanced (normal or translocation)? A.1/12 B. 1/6 C. 1/3 D. 1/2 E. 2/3

C. 1/3

The female Gryllus bimaculatus cricket has a diploid number of 28 chromosomes. How many chromosomes would you expect to find in the sperm cell of a male G. bimaculatus cricket? A.13 only B.14 only C.13 or 14 D.28 E.29

C. 13 or 14

The fruit fly (Drosophila melanogaster) is a model organism that has been studied by geneticists for over a century. The fruit fly genome has 4 pairs of chromosomes and crossing-over does not occur in male fruit flies. How many different combinations of chromosomes are possible following gametogenesis in male fruit flies? A.4 B.8 C.16 D.32 E.64

C. 16

In 2006, Hatakeyama et al. (Hum. Reprod. 21:976-979) submitted a case study of a male heterozygous for a Robertsonian translocation between chromosomes 13 and 21. They used FISH to analyze meiotic segregation patterns in the sperm of the study subject with the following results: Balanced normal or translocation 9036 Nullisomy 21 387 Disomy 21 208 Nullisomy 13 399 Disomy 13 139 As part of their study, Hatakeyama et al. reported the results of in vitro fertilization (IVF) and implantation using the man's sperm and ova from his wife. Assuming the data above are a representative sample of the man's sperm, what percentage of his children will be born with Down syndrome? Assume all trisomy 13's (Patau syndrome) survive. A. 1.4 B. 2.0 C. 2.2 D. 3.9 E. 5.9

C. 2.2

Suppose I want to have four children. What is the probability that I will have two girls and two boys? A. 6.25% B. 25% C. 37.5% D. 50% E. 66.6% F. 75%

C. 37.5%

Bread wheat (Triticum aestivum) is an allohexaploid (2n=6x=42) with two copies of each of three genomes: A, B and D. Durum wheat (Triticum turgidum; 2n=4x=28), the only domesticated tetraploid wheat in the world, consists of two copies each of the A and B genomes. If you cross bread wheat and durum wheat the progeny will be pentaploid. If you observe a cell from the pentaploid in metaphase I, what will you see (how will the chromosomes be arranged)? Specifically, how many mono-valents will you see? A.1 B. 6 C. 7 D. 14 E. 21 F. 28 G. 42 From which genome were they derived? A. A-genome B. B-genome C. D-genome

C. 7 C. D-genome

In 1916, Morgan and Bridges (Carnegie Inst. Publ. 327) reported the results of an experiment in which they crossed a wild-type, female D. melanogaster fruit fly with a male fly that had miniature wings and a sable body phenotype. All of the F1 progeny were wild type. They backcrossed female F1 flies with the parental male flies that had miniature wings and a sable body and obtained the following progeny. Wild type 793 Miniature, stable 668 Miniature 52 Sable 56 Are the alleles for miniature and sable in coupling or repulsion? What's the recombination frequency between the genes?

Coupling 0.0688

Bobbed (bb) is a pseudoautosomal recessive trait in Drosophila melanogaster that is characterized by shorter and thinner than normal bristles on the back of the head. If a bobbed male is mated with an attached X (X^XY) homozygous wild-type female, half of the adult F1 offspring are female and half are male but none are bobbed. What proportion of the adult F2 progeny will be bobbed? A.0 B.1/4 C.1/3 D.1/2 E.2/3

D. 1/2

The Manx breed of domesticated cats originated on the Isle of Man where they are common. They are a relatively old breed which is manifest by short, stubby tails. Various legends which explain why Manx cats have no tail includes Noah closing the door of the Ark and accidentally cutting off the Manx's tail. Another legend claims that Manx cats are the result of a cross between a cat and a rabbit—a so-called Cabbit. Manx cats are heterozygous for the tailless allele (T1T2), while homozygous tailless (T2T2) cats are spontaneously aborted before birth. If two Manx cats are allowed to mate, what proportion of their live-birth progeny will be tailless? A.1/16 B. 1/4 C. 1/2 D. 2/3 E. 3/4

D. 2/3

Common red clover, Trifolium pratense, is a diploid with 14 chromosomes per somatic cell. What is the somatic chromosome number of a triploid variant of this species? A. 7 B. 14 C. 15 D. 21 E. 28

D. 21

Parthenogenesis is a development pathway whereby unfertilized eggs develop into viable adult organisms. In the wasp, Bracon hebetor, parthenogenesis results in haploid organisms which are always phenotypically male. However, wasp offspring which arise from normal fertilization are almost always female. In 1939, P.W. Whiting (J. Morphology 66:323-355) showed that an X-linked gene with multiple alleles (Xa, Xb, Xc etc.) controls sex determination in Bracon hebetor: wasps that are homozygous or hemizygous at this X-linked locus are males, whereas wasps that are heterozygous are females. Suppose an Xa/Xb female is mated with a haploid Xa male such that 50 percent of the eggs she lays are fertilized by the male and 50 percent remain unfertilized. The unfertilized eggs will develop via parthenogenesis. What fraction of the offspring from this female's eggs will be males (including fertilized and unfertilized)? A.1/8 D. 3/

D. 3/4

Mikhaail-Philips et al. counted 558 sperm labeled with red on one end and green on the other end of chromosome 2, 160 sperm labeled with red on both ends of chromosome 2, and 185 sperm labeled with green on both ends of chromosome 2. What is the observed frequency of crossing over in the inverted region of chromosome 2? A.17.8% C. 19.1% E. 61.8% B.20.5% D. 38.2%

D. 38.2%

Mendel reported the results of a single dihybrid experiment in his paper. The experiment was carried out by first crossing a true-breeding plant which produced only yellow, round seeds when self-fertilized with a true-breeding plant which produced only green, wrinkled seeds when self-fertilized. All of the F1 progeny seed from this cross were yellow and round. Next, Mendel grew the F1 seeds and self-pollinated the plants to generate F2 progeny seeds. He observed in the progeny that 9/16 of the seeds were yellow and round, 3/16 were yellow and wrinkled, 3/16 were green and round, and 1/16 were green and wrinkled. How many different genotypes are represented among the yellow, round F2 progeny seeds? A.1 B. 2 C. 3 D. 4 E. 5 What fraction of all of the F2 plants (genotypes) were expected, when self fertilized, to produce only yellow seeds which would segregated for the round and wrinkled traits? A. 1/16 B. 1/8 C. 1

D. 4 B. 1/8

In 1916, Morgan and Bridges (Carnegie Inst. Publ. 327) reported the results of an experiment with D. melanogaster in which they crossed wild-type females with males that had yellow, sable bodies (yellow and sable are recessive traits). All of the F1 progeny were wild type. They allowed the F1 females and males to mate with one another to produce an F2 generation. The following lists male flies only in the F2 generation. wild type 101 yellow, sable 214 yellow 146 sable 96 Total: 557 Calculate the recombination frequency between yellow and sable and correct for double crossovers and interference using the Kosambi function. What is the linkage distance in cM between the two genes? A.35.4 B.43.4 C.44.1 D.66.4 E.70.2

D. 66.4

Pink-eye and albino are two recessive traits found in the deer mouse Peromyscus maniculatus. In mice with pink-eye (pp), the eye is devoid of color and appears pink from the blood vessels within it. Albino mice (aa) are completely lacking color both in their fur and in their eyes. F. H. Clark crossed pink-eyed mice (what this genotype?) with albino mice (what's this genotype?); the resulting F1 had normal coloration in their fur and eyes. He then crossed these F1 mice with mice that were pink eyed and albino (what's this genotype?) and obtained the following mice. Since it is very hard to distinguish between mice that are albino and mice that are both pink-eye and albino, he combined these two phenotypes into a single phenotype which is labelled as "albino, pink eye" (F. H. Clark. 1936. Journal of Heredity 27: 259-260). wild type fur and eye color 12 wild type fur, pink-eye 62 albino, pink eye 78 Total 152 What is

D. 76

Imagine that in blue-breasted fairy wrens the intensity of blue in the breast feathers is determined by four different loci (A, B, C and D), each with two alleles. The alleles act additively both within and between loci. At each locus, there exists an additive allele that produces blue pigment and a non-additive allele that does not. Assuming no environmental influences, how many phenotypic classes will there be for breast color? A. 3 B. 5 C. 7 D. 9 E. 11 What would be the frequency of the bluest-feather class in an F2 progeny population? A. 1/4 B. 1/16 C. 1/64 D. 1/256 E. 1/1024

D. 9 D. 1/256

Bread wheat (Triticum aestivum) is an allohexaploid with two copies of each of three genomes: A, B and D. Durum wheat (Triticum turgidum), the only domesticated tetraploid wheat in the world, consists of two copies each of the A and B genomes. If you were to cross bread wheat and durum wheat, what would the ploidy level of the progeny be? A. Diploid B. Triploid C. Tetraploid D. Pentaploid E. Hexaploid

D. Pentaploid

According to the entry for the disease in OMIM, hereditary hyperekplexia "is an early-onset neurologic disorder characterized by an exaggerated startle response to sudden, unexpected auditory or tactile stimuli. Affected individuals have brief episodes of intense, generalized hypertonia in response to stimulation. Neonates may have prolonged periods of rigidity and are at risk for sudden death from apnea or aspiration. Many affected infants have inguinal hernias. The symptoms tend to resolve after infancy, but adults may have increased startle-induced falls and/or experience nocturnal muscle jerks." Many cases of hyperekplexia are linked to mutations in the GLAR1 gene, which encodes the alpha subunit of the glycine receptor important to neurological function. In 2001, Rees et al. (Hum. Genet. 109:267-270) described a patient homozygous for a recessive GLAR1 allele named Tyr202Ter. What type of mutation resulted in t

D. Transversion mutation

Mendel self-pollinated his F2 monohybrid plants to test his hypothesis of a 2:1 ratio for heterozygotes to homozygotes. For example, he self-pollinated F2 purple-flowered plants and counted 64 that produced segregating progeny and 36 that produced non-segregating progeny. What is the probability of this particular outcome? A.1.847 x 10-16 B.2.636 x 10-10 C.0.0016 D.0.0042 E.0.0708

E. 0.0708

The ability of humans to taste the bitter compound phenylthiocarbamide (PTC) has long been thought to be controlled by a single gene called TASR38. Kim et al. (Science 299:1221-1225, 2003) described the recessive mutant allele A49P which prevents humans from tasting PTC. They reported the frequency of the mutant allele (q) as 0.47 among Europeans. Assuming these populations are in HWE, what proportion of Europeans can taste PTC? A.0.221 C. 0.498 E. 0.779 B.0.281 D. 0.719

E. 0.779

In 1991, Peter Koopman et al. reported the creation of transgenic mice in which they inserted the entire mouse Sry gene. They created the transgenic mice by injecting the SRY gene DNA into fertilized mouse eggs. They allowed 93 of the injected eggs to develop to full term. Five of the 93 mice tested positive for the presence of the transgenic SRY gene, including two XY male mice. If the injected SRY gene integrated on to one of the autosomes of these male (XY) mice and they can pass the transgene on, what would you expect the ratio of male to female mice to be in their progeny? A.1:1 D. 2:1 B.1:2 E. 3:1 C.1:3

E. 3:1

What is the expected phenotype of the F1 progeny kernels of a cross between a homozygous light color kernel plant, homozygous for the c1 allele and the Ac element (Ac+Ac+) that is crossed with a non-spotted plant homozygous for c-m1? Note that the c1 allele is due to a simple base mutation, whereas the c-m1 allele is due to the insertion of a Ds element. Both genes assort independently. A.All light-colored B.All purple-colored C.3:1 purple:light D.1:1 purple:spotted E.All spotted

E. All spotted

The maternal-effect mutation, bicoid (bcd), is recessive. In the absence of the bicoid protein product, embryogenesis is not completed. Consider a cross between a female heterozygous for the bicoid (bcd +/bcd -) and a male homozygous for the mutation (bcd -/bcd -). 1. What proportion of the F1 embryos will fail to develop? A. All embryos will fail to develop B. ¼ C. ½ D. ¾ E. None of the embryos will fail to develop 2. What proportion of the F2 embryos will fail to develop? A. All embryos will fail to develop B. ¼ C. ½ D. ¾ E.None of the embryos will fail to develop

E. None of the F1 embryo will fail to develop due to bicoid C. 1/2 will fail to develop

Leri-Weill Dyschondrosteosis (LWD; OMIM 127300) is a heritable skeletal dysplasia characterized by disproportionate short stature and mesomelic limb shortening. In 1998, Shears et al. (Nature Genet. 19:70-73) established that LWD is located in the pseudoautosomal region of the X and Y chromosomes. Further experiments by this research group led to the identification of mutations in the Short Stature Homeobox (SHOX) gene as responsible for the LWD mutant phenotype. In Family 7 they found a C-to-G mutation at nucleotide 597 of the SHOX coding region, converting a TAC (Tyr) codon to TAG (Ter). According to human gene nomenclature rules, what is the name of this allele? A.C597G B.G597C C.C199G D.Tyr597Ter E.Tyr199Ter

E. Try199Ter

In maize, kernels may either by purple or yellow. They may also be full or shrunken. For a particular experiment, a maize plant grown from a full, purple kernel was crossed with a plant grown from a shrunken, yellow kernel and the following progeny were counted on the ear obtained from the cross: full, purple 112 shrunken, purple 103 full, yellow 91 shrunken, yellow 94 Total 400 Which two traits are dominant? A.full and purple B.full and yellow C.shrunken and purple D.shrunken and yellow E.cannot determine

E. cannot determine

Mutations in the URF13 gene in maize (corn) mitochondria cause male sterility when the plant is homozygous for a recessive nuclear allele. In recent years, several nuclear genes have been discovered that are essential for restoration of male fertility in plants with a mutant URF13 allele. Two of these genes are named rf1 and rf2. (Recessive mutant alleles are designated in lower case, rf1 and rf2, and the dominant alleles as upper case Rf1 and Rf2.) These two nuclear genes assort independently of one another. For a plant to be male sterile, it must be homoplasmic for a mutant URF13 mitochondrial allele and homozygous for a recessive mutant allele in either the rf1 gene or the rf2 gene. Suppose that a male-sterile plant with the genotype rf1 rf1 / Rf2 Rf2 is hybridized with a male-fertile plant with the genotype Rf1 Rf1 / rf2 rf2. For the purposes of this question, assume that mitochondrial inheritance in maize is pu

F1 = URF13, rf1/Rf1 rf2/Rf2 (all fertile) F2 = 7/16 (all are ms mito)

In a cross of Limnaea, the snail contributing the eggs was dextral but of unknown genotype. Both the genotype and the phenotype of the other snail are unknown. All F1 offspring exhibited dextral coiling. Ten of the F1 snails were allowed to undergo self-fertilization. One-half produced only dextrally coiled offspring, whereas the other half produced only sinistrally coiled offspring. What were the genotypes of the original parents?

Mother is Rr and Father is rr

In 1920, blakeslee et al (Science 52:388-390) described the genetic and phenotypic effects of trisomy in jimsonweed. Plants with a normal diploid number of chromosomes (2n=24) have a normal phenotype, and plants that are trisomic for chromosome 12 (2n +1 = 25) have a phenotype called poinsettia. The p locus that governs flower color is on chromosome 12. Plants that carry the dominant P allele have purple flowers, and plants that are homozygous for the recessive p allele have white flowers. The authors of this study examined haploid pollen mother cells (the products of meiosis) from a poinsettia plant and found that half of the cells had one full set of n=12 chromosomes and the other half had n+1 =13 chromosomes, which is the expected segregation in a trisomic plant The authors also found that pollen grains with 13 chromosomes are inviable and cannot participate in fertilization, but pollen grains with n=12 chromosome

Ppp

In 1984, Martinez-Castro et al. (Cytogenetics and Cell Genetics 38:310-312) published a pedigree of a family in which two first cousins had married. They had six children, three of whom were homozygous for a Robertsonian rob(13q;14q) translocation and two who were heterozygous for a rob(13q;14q) translocation, and one with an unknown genotype who was stillborn (likely homozygous rob(13q;14q + 13; Patau syndrome). Assuming that the karyotypes of the parents are shown below, what is the probability of this outcome, i.e. a family of 6 live birth children with three homozygotes, two heterozygotes and one homozygous rob(13q;14q + 13)? Assume all trisomy 13's (Patau syndrome) survive.

Probability would be: [6!/3!2!1!](0.33)^3(0.33)^2(0.33)^1

Tatuo Aida investigated the genetic basis of color variation in the Medaka (Aplocheilus latipes), a small fish found naturally in Japan (T. Aida. 1921. Genetics 6:554-573). Aida found that genes at two loci (B, b and R, r) determine the color of the fish: fish with a dominant allele at bot loci (B_R_) are brown, fish with a dominant allele at the B locus only (B_rr) are blue, fish with a dominant allele at the R locus only (bbR_) are red, and fish with recessive alleles at both loci (bb rr) are white. Aida crossed a homozygous brown fish with a homozygous white fish. He then backcrossed the F1 with the homozygous white parent and obtained 228 brown fish, 230 blue fish, 237 red fish, and 222 white fish. What results would you expect for a cross between a homozygous red fish and a white fish?

Red (bbRr)

If a monosomic (2n-1,-4) female Drosophila (containing only one chromosome 4, but an otherwise normal set of chromosomes) that has white eyes (an X-linked trait) and normal bristles is crossed with a male with a diploid set of chromosomes and normal red eye color, but who is homozygous for the recessive chromosome 4 bristle mutation, shaven (sv), what phenotypic ratio do you expect in their progeny?

Red females with 1:1 ratio of shaven and normal bristles White males with 1:1 ratio of shaven and normal bristles

In peppered moths, a dominant allele D confers the dark phenotype when homozygous or heterozygous. The light colored moths are homozygous for a recessive allele d. In 1961, Kettlewell (Annual Review of Entomology 6:245-262) published his calculations of relative fitness values for light and dark moths in an area of Great Britain with polluted air. He marked equal numbers of light and dark moths and released them, then later determined the proportions of both types after recapturing the moths. He recaptured 53% of the dark-colored moths but only 25% of the light-colored moths. Assuming that the rate of recapture represent the actual frequencies of marked moths, (a) determine the relative fitness values for the light types and the dark types, (b) calculate the change in q after one generation of selection when q = 0.6.

Rf dark: 1 Rf light: 0.47 change in q = -0.09

A triply-heterozygous, wild-type plant in which the lo2 allele was in repulsion to the bz2 and wx alleles was crossed using pollen from a plant that was homozygous recessive for bz2 and wx and homozygous normal for Lo2. The phenotype of 150 viable kernels from the resultant ear were scored as indicated below. Bronze, waxy 114 Bronze 8 Waxy 27 Wild type 1 Total: 150 Why are there only four progeny types? What is the genetic linkage distance between bz2 and wx, in cM? A. 5.3 B. 6.0 C. 18.7 D. 24.0 E. 24.7

The other one's died E. 24.7

A particular cross between a tall pea plant with purple flowers and a tall pea plant with white flowers yielded progeny in the following proportions: 3/8 tall, purple 3/8 tall, white 1/8 dwarf, purple 1/8 dwarf, white (Assume independent assortment and that Tall and Purple are dominant) What are the genotypes of the two parents?

TtPp x Ttpp

In 1922, Hutchison reported the results of a testcross for alleles at three linked loci (c1, wx, and sh) on chromosome 9 in maize. One parent was colored, starchy and full and heterozygous at all three loci and the other parent was colorless, waxy and shrunken. Colorless, waxy and shrunken are the recessive phenotypes. •Test cross progeny: Colored, starchy, full (All wild type) 4 Colored, starchy, shrunken 2538 Colored, waxy, full 113 Colored, waxy, shrunken 601 Colorless, starchy, full 626 Colorless, starchy, shrunken 116 Colorless, waxy, full 2708 Colorless, waxy, shrunken 2 Total 6708 Determine the genetic map distances

Wx - Sh = 18.4 cM Sh - C1 = 3.5 cM

In 1929, Lambert and Knox (Poultry Science 9:51-64) reported the results of experiments on the inheritance of feathered and non-feathered shanks in chickens which they believe to be controlled by duplicate dominant epistatisis. In one set of experiments, they hybridized birds from a strain that bred true for feathered shanks (AABB) with birds from a strain that bred true for non-feathered shanks (aabb). All of the F1 offspring had feathered shanks. In the F2 generation, they observed 101 birds with feathered shanks and 10 with non-feathered shanks. Do there results conform to a duplicate dominant epistatic ratio (15:1)?

Yes

In Drosophila melanogaster, the vermilion phenotype is bright-red eyes and is the result of the recessive mutant allele v. The sable phenotype is black body color and is caused by the recessive allele s. The eosin phenotype is a reduction in both the brown and red pigments in the eyes and is caused by the recessive allele we. (Flies that are both eosin and vermilion can be readily distinguished from those that are either eosin or vermilion alone). All three genes are linked on the X chromosome. In 1916, Morgan and Bridges (Carnegie Inst. Publ. 327) reported the results of an experiment in which they crossed an eosin, sable female with a vermilion male. All of the F1 females were wild type and all of the F1 males were eosin, sable. They allowed the F1 females and males to mate with one another to produce an F2 generation. The following lists male flies only in the F2 generation. wild type 8 vermilion 291 sable 135 e

b. 0.1179; 0.4580 c. 0.6042

Uniparental-maternal heritance of mitochondria is also the typical mode in plants -Corn production in the US is reliant on the use of hybrid F1 corn -Seed corn producers have used mitochondrial male sterility to efficiently produce F1 seeds for many years Expression depends on interaction of alleles in the nuclear genome (Restorer gene [R/r]) and alleles in the mitochondria [ms vs. mf]) -Dominance (RR or Rr) at the Restorer (nuclear) gene allows for fertility, regardless of the mitochondria type. -Only when a plant is rr and ms (and mitochrodrial homoplasmic) is it male sterile. - -What is the phenotype of the following? -ms cytoplasm rr nuclear genotype -ms cytoplasm Rr nuclear genotype -ms cytoplasm RR nuclear genotype -mf cytoplasm rr nuclear genotype -mf cytoplasm Rr nuclear genotype -mf cytoplasm RR nuclear genotype

ms cytoplasm rr nuclear genotype

In 1985, R.M. May (Nature 315:12-13) calculated the relative fitness of the SS, SR, and RR genotypes in the presence of warfarin to be 0.68, 1.00 and 0.37, respectively. The reduced fitness of the RR genotype appears to result from an excessive requirement for vitamin K. Can you hypothesize as to why the most fit genotype is SR? •In the presence of warfarin, what is the value of the selection coefficient s for this population of rats? • •In the presence of warfarin, what is the value of the selection coefficient t for this population of rats? •If the frequency of the R allele (q0) in a population of Norway rats is initially 0.34, what will be its frequency (q1) after a single generation in the presence of warfarin?

s = 0.32 t = 0.63 0.339